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Why would a hydrogen atom "donate" to fluorine in an ionic bond but not in $\ce{HCl}$? Why would $\ce{H}$ and chlorine share instead of $\ce{Cl}$ just stripping it away like $\ce{F}$ does?

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Bonds can be completely covalent as in $\ce{Cl-Cl}$, $\ce{H-H}$, etc. In these cases the electron density is shared equally between the two atoms. Bonds can also be ionic as in $\ce{Na^{+} Cl^{-}}$ where much (~80%) of an electron has been transferred from the sodium atom to the chlorine atom. Between these two extreme cases exist a continuum of bonds that are called polar, covalent bonds. In such cases the electron density between the two atoms is not shared equally, but neither is the electron transferred extensively from one atom to the other.

Just as we can draw resonance structures for acetone and discuss how the real molecule is a weighted superposition of these two structures, we can draw resonance structures for inorganic compounds and, again, discuss how much the various structures count towards the "real" description of the molecule. In the diagram below, two resonance structures are shown for sodium chloride. The structure on the left represents a covalent bond and the structure on the right represents an ionic bond. Just as acetone is a mix of the two resonance structures, so is sodium chloride. When the weighted mix of the resonance structures lies far to the left, we can say the compound or bond is covalent; when it lies far to the right we can say the compound or bond is ionic. Everything in between represents a polar, covalent bond.

enter image description here

One can crudely estimate how polar a diatomic covalent bond is by using the following methodology. The dipole moment of a molecule is given by $$\ce{\mu~=~Q*d}$$ where the dipole moment is in Debye (D), Q is the partial charge on an atom measured in coulombs and d is the distance between the two atoms (the bond length) in meters.

In the case of $\ce{HF}$, the dipole moment is 1.86 D and the bond length is 91.7 x 10^(-12) m. Noting that 1 D = 3.3356 × 10^(−30) C·m, solving for Q we find a partial charge on one of the atoms of $$\mathrm{Q~=~\frac{[1.86 ~x~3.3356 ~x~ 10^{-30} ]}{[91.7~x~10^{-12}]}~=~6.77~x~10^{-20}~C}$$

Now since one electron equals 1.6022 × 10^(−19) C, by taking the ratio $$\mathrm{\frac{[6.77~x~10^{-20}]}{[1.6022 × 10^{−19}]}~=~ 0.42}$$ we estimate that roughly 42% of an electron has been transferred from the hydrogen atom to the fluorine atom. This gives us a rough idea of how ionic the polar, covalent bond is in $\ce{HF}$. Repeating this calculation for $\ce{HCl}$ suggests that roughly only 18% of an electron has been transferred in this case.

Clearly the polar, covalent bond in $\ce{HF}$ has more ionic character than the analogous polar, covalent bond in $\ce{HCl}$. The reason why the $\ce{HF}$ bond has more ionic character is the greater electronegativity of fluorine (3.98) compared to chlorine (3.16).

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  • $\begingroup$ Just to clarify, electron density can be donated and quantified in terms of percentages. $\endgroup$ – LordStryker Sep 4 '14 at 14:09

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